Is ${913231}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {913231}= &&{9}\cdot100000+ \\&&{1}\cdot10000+ \\&&{3}\cdot1000+ \\&&{2}\cdot100+ \\&&{3}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {913231}= &&{9}(99999+1)+ \\&&{1}(9999+1)+ \\&&{3}(999+1)+ \\&&{2}(99+1)+ \\&&{3}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {913231}= &&\gray{9\cdot99999}+ \\&&\gray{1\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{2\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {9}+{1}+{3}+{2}+{3}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${913231}$ is divisible by $3$ if ${ 9}+{1}+{3}+{2}+{3}+{1}$ is divisible by $3$ Add the digits of ${913231}$ $ {9}+{1}+{3}+{2}+{3}+{1} = {19} $ If ${19}$ is divisible by $3$ , then ${913231}$ must also be divisible by $3$ ${19}$ is not divisible by $3$, therefore ${913231}$ must not be divisible by $3$.